Pressure Depth in a Cave

[Scubakevdm]

This thread is absolutely shocking. Anyone who is convinced that the answer is 4 ATA should be flogged with his C-card.

The original poster's question is really just a convoluted version of the Hydrostatic Paradox.

You mustn't let looks deceive you...the pressure at point "C" has NOTHING to do with the solid rock above. It is defined solely by the height of the free water surface, and is therefore 2 ATA. You could replace the rock with marshmallows or pixie dust and the answer would be the same.

If you are still not convinced, consider the point 33ft down from the cave system entrance. The pressure there is certainly 2 ATA. If the pressure at C were anything other than 2 ATA there would be a pressure differential and the cave system would become an underwater river. For the system to be in equilibrium, point C MUST be at 2 ATA.

Class dismissed.

That's not entirely true, as pixie dust is water is both water soluable, and highly dense.

 

[Melicertes]

It is a system in hydrostatic equilibrium. Period.

I'm not even sure what you mean by "sealed system", nor how that has any bearing on this problem. Every version of your analysis so far is flawed.

Care to explain why?
I have already explained why I am arguing a closed/sealed (yes it's probably not the best wording for it) system model, and I've already stated that I realise that a cave in the real world cannot be considered a totally closed system. Read the posts further up first before just shooting the assertions down. Explaining how you get to your conclusion may also help me to see your point. "A system in hydrostatic equilibrium" doesn't tell me much - but then I'm no physicist so I'm not really surprised

 

[amascuba]

So, the correct answer at C is 2 ATA. And we know that C is not exposed to the atmosphere, otherwise there would be a net force at C (air would push down with 1 ATA and the water would push up with 2 ATA) so the water and air would move and water would push upwards until it reached a zero water column height (if the displacement of water caused the level of water at other areas in the system to drop, the water would rise until those heights were equal).

If you think about it that way wouldn't the answer technically be 1 ATA since you don't have the pressure of the atmosphere as a factor?

 

[Melicertes]

Maybe I should just bow out of this one and leave it be. I don't wish to aggrevate anyone. I'd love to learn through the discussion, but not at the expense of maintaining a friendly atmosphere of group participation.

 

[reefnet]

Care to explain why?

I just did in this post. There's no nice way to say the following, but realize it's nothing personal:

If you can't follow the article I linked to or the alternate explanation I offered, you probably shouldn't be diving. Furthermore, your status indicator suggests that you are an instructor. If that's true, then I would not hesitate to report your lack of understanding to your certifying agency.

That may sound hostile, but I'm honestly concerned for your safety. I'd hate for you (or a student) to get hurt because you made a mistake like this.

 

[Blackwood]

because I am arguing a perfect lab environment model as opposed to a real life cave system model and there are differences between the two which allows me to see a possible 2ATA answer to the real cave model and I maintain a 4ATA answer to the lab model. I'd love to continue the discussion, just know I'm not trying to trawl here

It has nothing to do with whether it's water in a lab, water in a cave or water in our imaginations...

 

[do it easy]

Something tells me that Lynne is posting as Peter Guy

 

[Benthic]

This thread is absolutely shocking. Anyone who is convinced that the answer is 4 ATA should be flogged with his C-card.

Agreed. This is elementary dive physics kids. If you don't get it--stay out of the water.

Brian

 

[ArcticDiver]

In your analysis consider the humble water level. It is a length of tubing with water in it; often many feet in length. Both ends are open to the atmosphere. One end is placed at the reference point and the other at the point where level is to be determined. Doesn't make any difference what happens to the intervening length as long as none of it is higher than the reference end.

 

[Blackwood]

I guess I am arguing this from an academic point of view and not so much a real life point of view. I have yet to see how others justify their 2ATA answer though if it's not for the reasons I have mentioned above.

If you REALLY want to open that can of worms, I suppose I’ve got nothing better to do than oblige.

Pressure is defined as the limit as ΔA->0 of ΔF/ΔA.

This yields the differential equation: p=dF/dA, where F is the normal force distributed over the area A.

Consider a fluid element ΔL long, ΔA in cross-sectional area, and inclined by the angle α to the horizontal. It is oriented such that its longitudinal axis is parallel to an arbitrary direction L.

The element is ΔL long, ΔA in cross-sectional area, and inclined by the angle α to the horizontal.

F=ma

We're (hydro)static, so the sum of all forces is zero. The appropriate forces to consider are pressure and that of gravity (weight).

ΣF=0

pΔA-(p+Δp)ΔA–ρΔAΔLsin(&#945=0, where ρ=density.

Simplify and divide by the volume of the fluid element (ΔAΔL)

Δp/ΔL=-ρsin(&#945

At the limit (ΔL->0), Δp/ΔL=dp/dL.

Also, sin(&#945=dz/dL, where z is vertical (i.e. the direction of gravity)

So dp/dL=-ρ(dz/dL), which means that a change of pressure in the L direction (i.e. dp/dL) ONLY occurs when there is a change of elevation.

C is at 2ATA.